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Sunday, 4 February 2018

STRENGTH OF MATERIALS Top 10 Numercial

Assume
g = 10 m/s2 for all questions

1. a. A round steel
bar with an initial diameter of 20 mm and length of 2 m is placed in tension, supporting
a load of 2000 kg. If the Young’s modulus of the bar is 200 GPa what is the
length of the bar when supporting the load (assuming it does not yield)?

b. If the Poisson’s
ratio of the steel is 0.4, what will the diameter of the bar be when supporting
the load?

Solution:

a. Cross-sectional
area = 3.14 × 10−4 m2

Load = 2000 kg × g = 2 × 104 N

Stress = 2 × 104/3.14 × 10−4 =
6.4 × 107 Pa

Strain = 6.4 × 107/200× 109 = 3.2 × 10−4

New length = (3.2 × 10−4× 2) + 2 = 2.00064 m

b. Lateral
strain = 3.2 × 10−4× 0.4 = 1.28 × 10−4

Change in diameter =
1.28 × 10−4× 0.02 = 2.56 × 10−6m

New diameter = 0.02 − 2.56 × 10−6= 1.999744 × 10−2m = 19.99744 mm

2. A 2 m long tie in a steel frame is made of
circular hollow section steel with an outer diameter of 100 mm and a wall
thickness of 5 mm. The properties of the steel are as follows:

Young’s
modulus: 200 GPa

Yield
stress: 300 MPa

Poisson’s ratio: 0.15

a. What is the extension of the tie when the
tensile load in it is 200 kN?

b.
What is the reduction in wall thickness when
this load of 200 kN is applied?

Solution:

a. Area
= π(0.052
− 0.0452)
= 1.491 ×10−3m2

Stress = 200 × 103/1.491 × 10−3=
1.34 × 108 Pa

This stress is 134
MPa, which is well below the yield stress.

Strain = 1.34 × 108/2 × 1011 = 6.7 × 10−4

Extension = 2 × 6.7 × 10−4=
0.00134 m = 1.34 mm

b. Strain
= 6.7 × 10−4× 0.15 = 1.005 × 10−4

Reduction = 1.005 × 10−4× 0.005 = 5.025 × 10−7m = 0.5 mm

3. A
water tank is supported by four identical timber posts that all carry an equal
load. Each post measures 50 mm by 75 mm in cross-section, and is 1.0 m long.
When 0.8 m3 of water is pumped into the tank, the posts get 0.07 mm shorter.

a. What
is the Young’s modulus of the timber in the direction of loading?

b. If
the cross-sections measure 75.002 mm by 50.00015 mm, after loading, what are
the relevant Poisson’s ratios?

c. If
300 L of water are now pumped out of the tank, what are the new dimensions of
the post?

(Assume that the
strain remains elastic.)

Solution:

Normally, all
calculations should be carried out in base units, but these calculations of
strain are an example of where it is clearly easier to work in mm.

a. 0.8
m3 of water has a mass of 800 kg and thus a weight of 800 × g =
8000 N

Stress = 8000/(4 × 0.075 × 0.05) = 5.33 × 105 Pa

Strain = 0.07/1000 =
7 × 10−5

Modulus = 5.33 × 105/7 × 10−5=
7.6 × 109 Pa

b. Change
in width on long side = 75.002 −
75 = 0.002 mm

Strain on long side =
0.002/75 = 2.66 × 10−5

Thus, Poisson ratio =
2.66 × 10−5/7 × 10−5=
0.38

Strain on short side
= 0.00015/50 = 3 × 10−6

Thus, Poisson ratio =
3 × 10−6/7 × 10−5=
0.043

c. 300
L of water has a mass of 300 kg.

Thus, the mass of
water now in the tank is 800 −
300 = 500 kg

The changes in
dimensions are proportional to the load, and thus they are 500/800 = 0.625 of those
in part (b).

0.07 × 0.625 = 0.044, thus length =
1000 − 0.044
= 999.956 mm

0.002 × 0.625 = 0.00125, thus long
side = 75.00125 mm

0.00015 × 0.625 =
0.000094, thus short side = 50.000094 mm

4. The
below figure shows observations that were made when a 100 mm length of 10 mm
diameter steel bar was loaded in tension. The equation given is for a trendline
that has been fitted to the straight portion of the graph. Calculate the
following:

a. the
Young’s modulus,

b. the
estimated yield stress,

c. the
ultimate stress.

Solution:

Cross-section area = π × 0.012/4 = 7.85 ×
10−5m2

a. The
equation given on the graph has been generated by the Excel compute package,
and is a best fit to the linear part of the data (this is real experimental data).
This technique is discussed in Chapter 15. The gradient taken from the equation
is 157.76 kN/mm = 1.5776 × 108 N/m

E =
1.5776 × 108 × 0.1/7.85 × 10−5=
2.01 × 1011
Pa = 201 GPa

b. Load
= 33 kN (estimated from the end of the linear section on the graph)

Thus, stress = 420
MPa

c. Load
= 42 kN at failure (the highest point on the graph)

Thus, stress = 535 MPa

5.
a. A 100 mm diameter concrete cylinder 200 mm long is loaded on the
end with 6 tonnes. What is the stress in it?

b. The
Young’s modulus of the cylinder is 25 GPa. What is its height after loading?

c. The
Poisson’s ratio of the cylinder is 0.17. What is its diameter after loading?

d. A
column 2 m high, measuring 300 mm by 500 mm on plan, is made with the same
concrete as in the cylinder and supports a bridge. In order to prevent damage
to the deck, the maximum permitted compression of the column is 0.3 mm.
Assuming that the column is not reinforced, calculate the compression of the
column when it is supporting the full weight of a 40 tonne lorry, and state
whether the deck will fail.

b. A
timber sample measuring 50 mm ×
50 mm on plan is supporting a mass of 2 tonnes. If

the relevant Young’s
modulus and Poisson’s ratio are 1.3 GPa and 0.4, what is the lateral

expansion caused by
the load?

Solution:

a. It
has 6 Poisson’s ratios (see Section 33.5.2):

longitudinal–tangential

tangential–longitudinal

radial–tangential

tangential–radial

radial–longitudinal

longitudinal–radial

b. Load
= 2 tonnes = 2000 kg

Weight = 2,000 × g =
20,000 N

Stress = 20,000/(0.05
× 0.05)
= 8 × 106 Pa

Longitudinal strain =
8 × 106/1.3
× 109 =
6.15 × 10−3

Lateral strain = 0.4 × 6.15 × 10−3 = 2.46 × 10−3

Expansion = 2.46 × 10−3× 0.05 = 1.23 × 10−4m = 0.123 mm

7. A
steel specimen measuring 1 in. by 0.5 in. in section is found to have yielded
at a load of 38 kips, and failed at 50 kips. Assuming that the modulus of
elasticity is 29,000 ksi, and the Poisson’s ratio is 0.3, calculate:

a. The
tensile stress at yield and failure.

b. The
strain when the stress is half the yield stress.

c. The
dimensions of the section at half the yield stress.

Solution:

a. Yield
stress = 38,000/(1 × 0.5)
= 76,000 psi

Failure stress =
50,000/(1 × 0.5)
= 105 psi

b. Longitudinal
strain = (0.5 × 76,000)/(2.9
× 107)
= 1.31 × 10−3

c. Lateral
strain = 1.31 × 10−3× 0.3 = 3.4 × 10−4

New width = 1 − (1 × 3.4 × 10−4) =
0.99966 in.

New thickness = 0.5 − (0.5 × 3.4 × 10−4) =
0.49983 in.

8.
a. A round steel bar with an initial diameter of ¾ in. and length of
6 ft. is placed in tension, supporting a load of 4,400 lb. If the Young’s
modulus of the bar is 29,000 ksi, what is the length of the bar when supporting
the load (assuming it does not yield)?

b. If
the Poisson’s ratio of the steel is 0.4, what will the diameter of the bar be
when supporting the load?

Solution:

a. Cross-sectional
area = 0.442 in.2

Stress = 4,400/0.442
= 9,954 psi

Strain = 9,954/29,000
× 103 =
3.43 ×10−4

New length = (3.43 × 10−4 × 6) + 6 = 6.002 ft.

b. Lateral
strain = 3.43 × 10−4× 0.4 = 1.37 × 10−4

Change in diameter =
1.37 × 10−4 × 0.75 = 1.027 × 10−4in.

New diameter = 0.75 − 1.027 × 10−4=
0.749897 in.

9. A 6
ft. long tie in a steel frame is made of circular hollow section steel with an
outer diameter of4 in., and a wall thickness of 0.2 in. The properties of the
steel are as follows:

Young’s modulus:
29,000 ksi

Yield stress: 43 ksi

Poisson’s ratio: 0.15

a. What
is the extension of the tie when the tensile load in it is 44 kips?

b. What
is the reduction in wall thickness when this load of 44 kips is applied?

Solution:

a. Area
= π(22 − 1.82) = 2.387 in.2

Stress = 44 × 103/2.387= 18.43 ksi

This stress is well
below the yield stress.

Strain = 18.43 × 103/2.9 × 107 = 6.35 × 10−4

Extension = 6 × 12 × 6.35 × 10−4 = 0.0457 in.

b. Strain
= 6.35 × 10−4 × 0.15 = 9.5 × 10−5

Reduction = 9.5 × 10−5 × 0.2 = 1.9 × 10−5in.

10. The
figure below shows observations which were made when a 4 in. length of 3/8 in.
diameter steel bar was loaded in tension. The equation given is for a trendline
that has been fitted to the straight portion of the graph. Calculate the
following:

a. the
Young’s modulus,

b. the
estimated yield stress,

c. the
ultimate stress.

Solution:

Cross-section area = π × 0.3752/4 = 0.11 in.2

a. The
equation given on the graph has been generated by the Excel compute package,
and is a best fit to the linear part of the data. The gradient taken from the
equation is 816.4 kips/in.

E = 816.4 × 4/0.11 = 29,687 ksi

b. Load
= 7.4 kips (estimated from the end of the linear section on the graph)